Does $\lfloor(4+\sqrt{11})^{n}\rfloor \pmod {100}$ repeat every $20$ cycles of $n$?

Solution 1:

Consider the linear recurrence relation $a_{n+2} = 8a_{n+1} - 5a_n$ with $a_0 = 2$ and $a_1 = 8$. It is clear that $a_n$ is an integer for all $n \geq 0$. Also, this recurrence has the following closed-form solution for all $n \geq 0$:

$a_n = (4+ \sqrt{11})^n + (4 - \sqrt{11})^n$

Since $0 < 4 - \sqrt{11} < 1$, it follows that $\lfloor(4+ \sqrt{11})^n\rfloor = a_n - 1$ for all $n \geq 0$. Periodicity of $\lfloor(4+ \sqrt{11})^n\rfloor$ follows almost immediately now: first we check that $54 = a_{2} \equiv a_{22} \text{ mod } 100$ [Edit: and $92 \equiv a_{3} \equiv a_{23} \text{ mod } 100$]. Since $a_n$ is defined via a linear recurrence, it follows that $a_n \equiv a_{n+20} \text{ mod } 100$ for all $n \geq 2$, so $\lfloor(4+ \sqrt{11})^n\rfloor \equiv \lfloor(4+ \sqrt{11})^{n+20}\rfloor \text{ mod } 100$ for all $n \geq 2$ as well.